Exercises — Gravity turn trajectory — pitch rate from aerodynamic angle of attack = 0

The turning is done by gravity — specifically its component perpendicular to the velocity vector. Gravity has magnitude $mg$ pointing straight down. Split it relative to the velocity (which sits at angle $\gamma$ above horizon): the perpendicular slice has size $mg\cos \gamma .$ Thrust and drag are tangential (because $\alpha =0$), so they add nothing sideways. Look at figure s01: the red arrow (labelled $mg\cos \gamma$) is the only sideways pull.

$\dot{\gamma }=0$. Vertical means $\gamma =90^{\circ }$, and $\cos 90^{\circ }=0$, so the perpendicular gravity component vanishes. Gravity points straight back down the velocity line — it can slow the rocket but has no sideways grip to turn it. A vertical rocket stays vertical forever.

The minus sign says $\gamma$ is decreasing: the velocity arrow tips down toward the horizon over time. That is exactly what "turning over" means — the rocket leans from vertical toward horizontal. If the sign were $+$, the rocket would rear up, which gravity never does.

Plug into the boxed law. $\cos 60^{\circ }=0.5$. $\dot{\gamma }=-\frac{9.81\times 0.5}{80}=-\frac{4.905}{80}=-0.06131\text{rad/s}.$ Convert: multiply by $180/\pi \approx 57.30$: $-0.06131\times 57.30\approx -3.51^{\circ }/\text{s}.$ Reading it: the nose is tipping over at about three and a half degrees each second — a brisk turn because $\gamma$ is well below vertical and $v$ is still small.

Rearrange the law for $v$. Start from $|\dot{\gamma }|=\frac{g\cos \gamma }{v}\Rightarrow v=\frac{g\cos \gamma }{|\dot{\gamma }|}.$ With $\cos 50^{\circ }=0.6428$: $v=\frac{9.81\times 0.6428}{0.02}=\frac{6.306}{0.02}=315.3\text{m/s}.$ Why solve for $v$? We were given the desired rate and asked for the condition that produces it — so $v$ is the unknown, and the algebra is one clean division.

Same formula, just swap $g$. $\cos 30^{\circ }=0.8660$. $\dot{\gamma }=-\frac{1.62\times 0.8660}{40}=-\frac{1.4029}{40}=-0.03507\text{rad/s}.$ In degrees: $-0.03507\times 57.30\approx -2.01^{\circ }/\text{s}$. Lesson: weaker gravity ⇒ weaker sideways pull ⇒ slower gravity turn. On the Moon you must lean over more deliberately because gravity helps less.

Since $\gamma$ (and hence $\cos \gamma$) and $g$ are identical, only $1/v$ differs. So $\frac{{\dot{\gamma }}_A}{{\dot{\gamma }}_B}=\frac{1/v_A}{1/v_B}=\frac{v_B}{v_A}=\frac{360}{120}=3.$ Rocket A turns three times faster. Numeric check: $\cos 45^{\circ }=0.7071$. ${\dot{\gamma }}_A=-9.81(0.7071)/120=-0.05781$ rad/s; ${\dot{\gamma }}_B=-9.81(0.7071)/360=-0.01927$ rad/s; ratio $=3.00$. ✅ Insight: speed is a brake on turning. The slow rocket bends its path far more per second — which is exactly why the critical pitch-over is done early, while $v$ is small.

At $\gamma =0$: $|{\dot{\gamma }}_0|=g\cos 0/v=g/v$ (since $\cos 0=1$). We want $\frac{g\cos \gamma }{v}=\frac{1}{2}\cdot \frac{g}{v}\Rightarrow \cos \gamma =\frac{1}{2}.$ The $g$ and $v$ cancel — so the answer is a pure angle: $\gamma =\arccos (0.5)=60^{\circ }$. Why $\arccos$? We know the cosine ($0.5$) and want the angle. $\arccos$ is the tool that undoes cosine — it answers "which angle has this cosine?" The unique answer in $[0^{\circ },90^{\circ }]$ is $60^{\circ }$.

Why $r=v/\dot{\gamma }$? In curved motion the normal (centre-pointing) acceleration is $a_{\perp }=v\dot{\gamma }$, and also $a_{\perp }=v^2/r$ (see Centripetal / Normal Acceleration). Setting them equal: $v^2/r=v\dot{\gamma }\Rightarrow r=v/\dot{\gamma }$. Now substitute the gravity-turn law $|\dot{\gamma }|=g\cos \gamma /v$: $r=\frac{v}{|\dot{\gamma }|}=\frac{v}{g\cos \gamma /v}=\frac{v^2}{g\cos \gamma }.$ Numbers: $\cos 40^{\circ }=0.7660$. $r=\frac{150^2}{9.81\times 0.7660}=\frac{22500}{7.515}\approx 2994\text{m}\approx 3.0\text{km}.$ Reading it: the path curves along a circle of radius $\sim 3$ km at that instant. As $v$ grows, $r$ grows like $v^2$ — the path straightens out fast.

Step 1 — separate. Group all $\gamma$ on one side, $t$ on the other: $\frac{d\gamma }{\cos \gamma }=-\frac{g}{\bar{v}}dt.$ Why? Time doesn't appear on the left except through $\gamma$, so isolating $\gamma$ lets each side be integrated on its own. Step 2 — where does $\int \frac{d\gamma }{\cos \gamma }=\ln |\sec \gamma +\tan \gamma |$ come from? This isn't magic — here is the one-line reason. Multiply top and bottom of $\sec \gamma =1/\cos \gamma$ by the clever factor $(\sec \gamma +\tan \gamma )$: $\sec \gamma =\sec \gamma \cdot \frac{\sec \gamma +\tan \gamma }{\sec \gamma +\tan \gamma }=\frac{{\sec }^2\gamma +\sec \gamma \tan \gamma }{\sec \gamma +\tan \gamma }.$ Now look at the denominator $u=\sec \gamma +\tan \gamma$. Its derivative is $\frac{du}{d\gamma }=\sec \gamma \tan \gamma +{\sec }^2\gamma$ — which is exactly the numerator! So the integrand is of the form $u^{\prime }/u$, whose integral is $\ln |u|$: $\int \sec \gamma d\gamma =\int \frac{u^{\prime }}{u}d\gamma =\ln |u|=\ln |\sec \gamma +\tan \gamma |.$ (Geometrically, $\sec \gamma +\tan \gamma$ is the length swept along the tangent line as $\gamma$ opens up — the log appears because that length grows multiplicatively, the same way $\ln$ shows up whenever a quantity grows in proportion to itself.) Step 3 — apply the bounds. $[\ln |\sec \gamma +\tan \gamma |{]}_{{\gamma }_1}^{{\gamma }_2}=-\frac{g}{\bar{v}}\Delta t.$ At ${\gamma }_2=70^{\circ }$: $\sec 70^{\circ }=2.924$, $\tan 70^{\circ }=2.747$, sum $=5.671$, $\ln =1.7354$. At ${\gamma }_1=80^{\circ }$: $\sec 80^{\circ }=5.759$, $\tan 80^{\circ }=5.671$, sum $=11.430$, $\ln =2.4362$. Left side $=1.7354-2.4362=-0.7008$. Step 4 — solve for $\Delta t$. $-0.7008=-\frac{9.81}{200}\Delta t=-0.04905\Delta t\Rightarrow \Delta t=\frac{0.7008}{0.04905}\approx 14.29\text{s}.$ Sanity check: near vertical the turn is slow (small $\cos \gamma$), so tipping just $10^{\circ }$ takes a good chunk of time — about 14 s is reasonable.

$\gamma$ (hence $\cos \gamma$) and $g$ are fixed, so $|\dot{\gamma }|\propto 1/v$. The ratio end-to-start: $\frac{|{\dot{\gamma }}_{\text{end}}|}{|{\dot{\gamma }}_{\text{start}}|}=\frac{v_{\text{start}}}{v_{\text{end}}}=\frac{100}{400}=\mathrm{0.25.}$ The turn rate drops to a quarter as the engine accelerates the vehicle. Guidance meaning: almost all the shaping of the trajectory happens while the rocket is slow. By the time it's fast, gravity barely bends the path, so the pitch program is essentially "set the initial lean correctly and then let physics coast." This is why a small pitch-over error early is far more costly than a late one.

As $\gamma \to 0$, $\cos \gamma \to 1$ (its largest value), so the per-radian pull is strongest there — that helps. But two things fight it:

1. In a real ascent $v$ grows large (thousands of m/s), and $|\dot{\gamma }|=g\cos \gamma /v\to$ small because $v$ is huge. So the turn rate shrinks even though $\cos \gamma$ is near its max.
2. In the flat-Earth model gravity always has a downward component pulling $\gamma$ below $0^{\circ }$ once horizontal, so pure gravity would keep tipping the nose under the horizon — you'd start descending. Conclusion: a pure flat-Earth gravity turn can pass through $\gamma =0$, but it doesn't stay there — gravity keeps pulling $\gamma$ negative. Real orbital insertion needs Earth's curvature: as the rocket goes fast enough, the ground curves away beneath it and "level" is a moving target, letting $\gamma$ settle near $0$ without diving. The flat-Earth law is an early-ascent tool only; near burnout you must switch to a curved-Earth model.

The constraint is $|\dot{\gamma }|\le 0.05$, i.e. $\frac{g\cos \gamma }{v}\le 0.05\Rightarrow v\ge \frac{g\cos \gamma }{0.05}.$ With $\cos 85^{\circ }=0.08716$: $v_{\min }=\frac{9.81\times 0.08716}{0.05}=\frac{0.8550}{0.05}=17.1\text{m/s}.$ So any speed above $\sim 17\text{m/s}$ already satisfies the limit at $\gamma =85^{\circ }$ — comfortably easy, because near-vertical flight has tiny $\cos \gamma$. The real danger comes later, not now. As $\gamma$ falls toward level, $\cos \gamma$ climbs toward $1$, so the required minimum speed climbs with it. At the extreme horizontal case $\gamma =0^{\circ }$: $v_{\min }(\gamma =0)=\frac{9.81\times \cos 0}{0.05}=\frac{9.81}{0.05}=196.2\text{m/s}.$ So the constraint is loosest near vertical ($17\text{m/s}$ is plenty) but tightest near horizontal ($196\text{m/s}$ needed). A rocket that is still slower than $\sim 196\text{m/s}$ when its path reaches level would exceed the payload's angular-rate limit and could damage it. Design fix: ensure the vehicle has accelerated past $\approx 196\text{m/s}$ before $\gamma$ approaches $0$ — i.e. gain speed early so the late, low-$\gamma$ part of the turn stays gentle. Launching too slowly is dangerous precisely because the singular-worst moment is the low-$\gamma$ end, not the loud, slow lift-off.

Speed from the constraint (as in L5.2), $\cos 40^{\circ }=0.7660$: $v=\frac{g\cos \gamma }{0.05}=\frac{9.81\times 0.7660}{0.05}=\frac{7.515}{0.05}=150.3\text{m/s}.$ Radius from L3.3's formula (or directly $r=v/|\dot{\gamma }|$): $r=\frac{v}{|\dot{\gamma }|}=\frac{150.3}{0.05}=3006\text{m}\approx 3.0\text{km}.$ (Check via $r=v^2/(g\cos \gamma )=150.3^2/7.515=3006$ m. ✅) Reading it: enforcing a max turn rate fixes both how fast you must be going and how gently the path curves — one payload spec locks two trajectory numbers at once. Notice $r=v/|\dot{\gamma }|$ is the cleanest route: at the constraint limit, radius is just speed divided by the allowed rate.

• Which force turns the rocket? ::: gravity's perpendicular component $mg\cos \gamma$.
• Turn rate vs speed? ::: $|\dot{\gamma }|\propto 1/v$ — faster = slower turn.
• Turn radius formula? ::: $r=v^2/(g\cos \gamma )$.
• Antiderivative of $\sec \gamma$? ::: $\ln |\sec \gamma +\tan \gamma |$.
• Where does a turn-rate limit bite hardest? ::: at the smallest $\gamma$ (near horizontal), where $\cos \gamma \to 1$.
• What breaks the law at liftoff? ::: $v\to 0$ makes $1/v$ diverge, and $\gamma =90^{\circ },v=0$ is a $0/0$ indeterminate — so we seed a pitch-over once $v>0$.